Metamath Proof Explorer

Theorem grpinvid1

Description: The inverse of a group element expressed in terms of the identity element. (Contributed by NM, 24-Aug-2011)

Ref Expression
Hypotheses grpinv.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
grpinv.p ⊢ + = ( +g ‘ 𝐺 )
grpinv.u ⊢ 0 = ( 0g ‘ 𝐺 )
grpinv.n ⊢ 𝑁 = ( invg ‘ 𝐺 )
Assertion grpinvid1 ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑁 ‘ 𝑋 ) = 𝑌 ↔ ( 𝑋 + 𝑌 ) = 0 ) )

Proof

Step Hyp Ref Expression
1 grpinv.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
2 grpinv.p ⊢ + = ( +g ‘ 𝐺 )
3 grpinv.u ⊢ 0 = ( 0g ‘ 𝐺 )
4 grpinv.n ⊢ 𝑁 = ( invg ‘ 𝐺 )
5 oveq2 ⊢ ( ( 𝑁 ‘ 𝑋 ) = 𝑌 → ( 𝑋 + ( 𝑁 ‘ 𝑋 ) ) = ( 𝑋 + 𝑌 ) )
6 5 adantl ⊢ ( ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ∧ ( 𝑁 ‘ 𝑋 ) = 𝑌 ) → ( 𝑋 + ( 𝑁 ‘ 𝑋 ) ) = ( 𝑋 + 𝑌 ) )
7 1 2 3 4 grprinv ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( 𝑋 + ( 𝑁 ‘ 𝑋 ) ) = 0 )
8 7 3adant3 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 + ( 𝑁 ‘ 𝑋 ) ) = 0 )
9 8 adantr ⊢ ( ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ∧ ( 𝑁 ‘ 𝑋 ) = 𝑌 ) → ( 𝑋 + ( 𝑁 ‘ 𝑋 ) ) = 0 )
10 6 9 eqtr3d ⊢ ( ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ∧ ( 𝑁 ‘ 𝑋 ) = 𝑌 ) → ( 𝑋 + 𝑌 ) = 0 )
11 oveq2 ⊢ ( ( 𝑋 + 𝑌 ) = 0 → ( ( 𝑁 ‘ 𝑋 ) + ( 𝑋 + 𝑌 ) ) = ( ( 𝑁 ‘ 𝑋 ) + 0 ) )
12 11 adantl ⊢ ( ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ∧ ( 𝑋 + 𝑌 ) = 0 ) → ( ( 𝑁 ‘ 𝑋 ) + ( 𝑋 + 𝑌 ) ) = ( ( 𝑁 ‘ 𝑋 ) + 0 ) )
13 1 2 3 4 grplinv ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( ( 𝑁 ‘ 𝑋 ) + 𝑋 ) = 0 )
14 13 oveq1d ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( ( ( 𝑁 ‘ 𝑋 ) + 𝑋 ) + 𝑌 ) = ( 0 + 𝑌 ) )
15 14 3adant3 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( ( 𝑁 ‘ 𝑋 ) + 𝑋 ) + 𝑌 ) = ( 0 + 𝑌 ) )
16 1 4 grpinvcl ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( 𝑁 ‘ 𝑋 ) ∈ 𝐵 )
17 16 adantrr ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → ( 𝑁 ‘ 𝑋 ) ∈ 𝐵 )
18 simprl ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → 𝑋 ∈ 𝐵 )
19 simprr ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → 𝑌 ∈ 𝐵 )
20 17 18 19 3jca ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → ( ( 𝑁 ‘ 𝑋 ) ∈ 𝐵 ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) )
21 1 2 grpass ⊢ ( ( 𝐺 ∈ Grp ∧ ( ( 𝑁 ‘ 𝑋 ) ∈ 𝐵 ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → ( ( ( 𝑁 ‘ 𝑋 ) + 𝑋 ) + 𝑌 ) = ( ( 𝑁 ‘ 𝑋 ) + ( 𝑋 + 𝑌 ) ) )
22 20 21 syldan ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → ( ( ( 𝑁 ‘ 𝑋 ) + 𝑋 ) + 𝑌 ) = ( ( 𝑁 ‘ 𝑋 ) + ( 𝑋 + 𝑌 ) ) )
23 22 3impb ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( ( 𝑁 ‘ 𝑋 ) + 𝑋 ) + 𝑌 ) = ( ( 𝑁 ‘ 𝑋 ) + ( 𝑋 + 𝑌 ) ) )
24 15 23 eqtr3d ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 0 + 𝑌 ) = ( ( 𝑁 ‘ 𝑋 ) + ( 𝑋 + 𝑌 ) ) )
25 1 2 3 grplid ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵 ) → ( 0 + 𝑌 ) = 𝑌 )
26 25 3adant2 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 0 + 𝑌 ) = 𝑌 )
27 24 26 eqtr3d ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑁 ‘ 𝑋 ) + ( 𝑋 + 𝑌 ) ) = 𝑌 )
28 27 adantr ⊢ ( ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ∧ ( 𝑋 + 𝑌 ) = 0 ) → ( ( 𝑁 ‘ 𝑋 ) + ( 𝑋 + 𝑌 ) ) = 𝑌 )
29 1 2 3 grprid ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑁 ‘ 𝑋 ) ∈ 𝐵 ) → ( ( 𝑁 ‘ 𝑋 ) + 0 ) = ( 𝑁 ‘ 𝑋 ) )
30 16 29 syldan ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( ( 𝑁 ‘ 𝑋 ) + 0 ) = ( 𝑁 ‘ 𝑋 ) )
31 30 3adant3 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑁 ‘ 𝑋 ) + 0 ) = ( 𝑁 ‘ 𝑋 ) )
32 31 adantr ⊢ ( ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ∧ ( 𝑋 + 𝑌 ) = 0 ) → ( ( 𝑁 ‘ 𝑋 ) + 0 ) = ( 𝑁 ‘ 𝑋 ) )
33 12 28 32 3eqtr3rd ⊢ ( ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ∧ ( 𝑋 + 𝑌 ) = 0 ) → ( 𝑁 ‘ 𝑋 ) = 𝑌 )
34 10 33 impbida ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑁 ‘ 𝑋 ) = 𝑌 ↔ ( 𝑋 + 𝑌 ) = 0 ) )